A water heater is fueled by the combustion of propane as described by the following reaction:
C3H8(g) + 5 O2(g) --> 3 CO2(g) + 4 H2O(l)
a) Determine how much heat is released if 11.2 moles of propane are burned at 298 K and at a constant pressure of 1.00 bar.
b) Assume that the water heater contains 146 L of water at an initial temperature of 25.0 degreeC. What will the water temperature rise to if all of the heat produced under the previous conditions is used to heat the water?
ANS:
a) -2.49x10^7 J
b) 65.8 degreeC
Please explain steps. Thank you! Much help is appreciated.
C3H8 + 5O2 ==> 3CO2 + 4H2O(l)
dHrxn = (n*dHf products) - (n*dHf reactants) and that x 11.2 for total dH.
b.
dH from above(+ sign here) = mass H2O x specific heat H2O x (Tfinal-Tinitial). Solve for Tf. I worked the problem and obtained those answers. By the way, thanks for including them in the post. I wish others would do that; it makes sure I don't drop a sign or forget a conversion.
Oh no problem :) and thank you so much for answering!
To solve these problems, we will use the concept of stoichiometry and the enthalpy change of the reaction.
a) To calculate the heat released when 11.2 moles of propane are burned, we will first calculate the number of moles of CO2 produced by using the stoichiometry of the reaction.
From the balanced equation, we can see that 1 mole of C3H8 produces 3 moles of CO2. So, the number of moles of CO2 produced will be:
11.2 moles C3H8 x (3 moles CO2 / 1 mole C3H8) = 33.6 moles CO2
Now, we can use the balanced equation and the enthalpy change (∆H) of the reaction to calculate the heat released. The enthalpy change for the given reaction is not provided, but we can assume that it is -2220 kJ/mol (standard enthalpy of combustion for propane).
The heat released can be calculated using the equation:
Heat released = ∆H x moles of CO2 produced
So, the heat released will be:
Heat released = -2220 kJ/mol x 33.6 moles = -74,592 kJ
We need to convert this value to joules since the answer is needed in joules:
1 kJ = 1000 J
Therefore, the heat released will be:
-74,592 kJ x 1000 J/kJ = -7.46 x 10^7 J
However, we are given that the reaction occurs at a constant pressure of 1.00 bar. So the heat released is at constant pressure, which is also equal to the change in enthalpy (∆H) of the reaction. Thus, the heat released is also equal to the ∆H of the reaction, -7.46 x 10^7 J.
Therefore, the heat released if 11.2 moles of propane are burned is -2.49 x 10^7 J (rounded to 3 significant figures).
b) To determine the change in water temperature, we will use the equation:
q = mcΔT
Where q is the heat transferred (in joules), m is the mass of water (in kilograms), c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (in °C).
First, we need to calculate the mass of the water. We know that 1 liter of water weighs 1 kg, so 146 L of water weighs 146 kg.
Next, we substitute the known values into the equation and solve for ΔT:
-2.49 x 10^7 J = (146 kg) x (4.18 J/g°C) x ΔT
Now, let's solve for ΔT:
ΔT = -2.49 x 10^7 J / (146 kg x 4.18 J/g°C)
ΔT ≈ 65.8 °C
Therefore, the water temperature will rise to approximately 65.8 °C when all the heat produced is used to heat the water.
To solve these problems, we need to use the concept of heat transfer and the principles of stoichiometry.
a) To determine the amount of heat released when burning 11.2 moles of propane at 298 K and 1.00 bar pressure, we first need to calculate the number of moles of propane that participate in the combustion reaction.
From the balanced equation, we can see that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2) to produce 4 moles of water (H2O). Therefore, the mole ratio of propane to water is 1:4.
Since we have 11.2 moles of propane, we can calculate the moles of water produced by multiplying it by the mole ratio:
11.2 moles C3H8 * (4 moles H2O / 1 mole C3H8) = 44.8 moles H2O
Now, we need to determine the amount of heat released per mole of propane burned. This value is typically known as the molar heat of combustion (ΔHc) of propane and is given as -2220 kJ/mol (negative sign indicates heat release during the reaction).
To convert the ΔHc from kJ/mol to J/mol, we multiply by 1000:
-2220 kJ/mol * 1000 J/kJ = -2.22 × 10^6 J/mol
Finally, we can calculate the total heat released by multiplying the moles of propane burned by the molar heat of combustion:
44.8 moles H2O * -2.22 × 10^6 J/mol = -9.94 × 10^7 J
Therefore, the amount of heat released is -9.94 × 10^7 J (or approximately -2.49 × 10^7 J after rounding to two significant figures).
b) To determine the increase in water temperature, we need to consider the heat gained by the water. The heat gained by the water is equal to the heat released during combustion.
First, we need to convert the initial water volume from liters (L) to kilograms (kg). Since the density of water is 1 g/cm^3 or 1000 kg/m^3, the mass can be calculated as follows:
146 L * 1 kg/L = 146 kg
Next, we need to calculate the specific heat capacity of water (Cw). The specific heat capacity of water is 4.18 J/g°C or 4180 J/kg°C.
Now we can use the equation:
Heat gained = mass of water * specific heat capacity * change in temperature (ΔT)
Rearranging the equation, we can solve for the change in temperature:
ΔT = Heat gained / (mass of water * specific heat capacity)
Plugging in the values:
ΔT = (-9.94 × 10^7 J) / (146 kg * 4180 J/kg°C)
ΔT ≈ 65.8 °C (after rounding to two significant figures)
Therefore, the water temperature will rise to approximately 65.8°C after all the heat produced is used to heat the water.